Question

In a certain​ country, the true probability of a baby being a boy is 0.518. Among the next four randomly selected births in the​ country, what is the probability that at least one of them is a girl​? ​(Round to three decimal places as​ needed.)

Answer 1

Answer:

0.928 = 92.8% probability that at least one of them is a girl

Step-by-step explanation:

For each baby, there are only two possible outcomes. Either they are boys, or they are not. The probabilities of each baby being a boy is independent from other babies. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

In a certain​ country, the true probability of a baby being a boy is 0.518. So $p = 0.518$

Among the next four randomly selected births in the​ country, what is the probability that at least one of them is a girl​?

Either all four babies are boys, or at least one is girl. The sum of these events is decimal 1. So

$P(X = 4) + P(X < 4) = 1$

We want to find $P(X < 4)$ when $n = 4$. So

$P(X < 4) = 1 - P(X = 4)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{4,4}.(0.518)^{4}.(0.482)^{0} = 0.072$

$P(X < 4) = 1 - P(X = 4) = 1 - 0.072 = 0.928$

0.928 = 92.8% probability that at least one of them is a girl